6161103 2.4 addition of a system

2.4 Addition of a System of Coplanar Forces


For resultant of two or more forces:� Find the components of the forces in the

specified axes

� Add them algebraically

� Form the resultant

In this subject, we resolve each force into rectangular forces along the x and y axes.

yx FFF +=



� Scalar Notation- x and y axes are designated positive and negative

- Components of forces expressed as algebraic - Components of forces expressed as algebraic scalars

Eg:

Sense of direction

along positive x and

y axes

yx FFF +=



�Scalar NotationEg:

Sense of direction

yx FFF ''' +=

Sense of direction

along positive x and

negative y axes



�Scalar Notation- Head of a vector arrow = sense of the vector graphically (algebraic signs not vector graphically (algebraic signs not used)

- Vectors are designated using boldface notations

- Magnitudes (always a positive quantity) are designated using italic symbols



� Cartesian Vector Notation- Cartesian unit vectors i and j are used to designate the x and y directions

- Unit vectors i and j have dimensionless - Unit vectors i and j have dimensionless magnitude of unity ( = 1 )

- Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis)

- Magnitude is always a positive quantity, represented by scalars Fx and Fy



� Cartesian Vector Notation F = Fxi + Fyj F’ = F’xi + F’y(-j)

F’ = F’xi – F’yj



� Coplanar Force ResultantsTo determine resultant of several coplanar forces:

- Resolve force into x and y - Resolve force into x and y components

- Addition of the respective components using scalar algebra

- Resultant force is found using the parallelogram law



� Coplanar Force ResultantsExample: Consider three coplanar forces

Cartesian vector notation

F1 = F1xi + F1yj

F2 = - F2xi + F2yj

F3 = F3xi – F3yj

� Coplanar Force ResultantsVector resultant is therefore

FR = F1 + F2 + F3

= F i + F j - F i + F j + F i – F j



= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj

= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j

= (FRx)i + (FRy)j



� Coplanar Force ResultantsIf scalar notation are usedFRx = (F1x - F2x + F3x)

FRy = (F1y + F2y – F3y)FRy = (F1y + F2y – F3y)

In all cases,

FRx = ∑Fx

FRy = ∑Fy

* Take note of sign conventions



� Coplanar Force Resultants- Positive scalars = sense of direction along the positive coordinate axes

- Negative scalars = sense of direction - Negative scalars = sense of direction along the negative coordinate axes

- Magnitude of FR can be found by Pythagorean Theorem

RyRxR FFF 22 +=



� Coplanar Force Resultants- Direction angle θ (orientation of the force) can be found by trigonometry

Rx

Ry

F

F1tan−=θ



Example 2.5

Determine x and y components of F1 and F2

acting on the boom. Express each force as a

Cartesian vector Cartesian vector



Solution

Scalar Notation

←=−=−= NNNF x 10010030sin2001o

Hence, from the slope triangle

↑=== NNNF y

x

17317330cos2001

1

o

= −

12

5tan1θ



Solution

Alt, by similar triangles

N

F x

13

12

2602 =

Similarly,

NNF

N

x 24013

12260

13260

2 =

=

=

NNF y 10013

52602 =

=



Solution

Scalar Notation

→== NNF x 2402402

Cartesian Vector Notation

F1 = {-100i +173j }N

F2 = {240i -100j }N

↓=−=

→==

NNF

NNF

y

x

100100

240240

2

2



Example 2.6

The link is subjected to two forces F1 and

F2. Determine the magnitude and F2. Determine the magnitude and orientation of the resultant force.



Solution

Scalar Notation

Σ= FF xRx :

↑=

+=

Σ=

→=

−=

N

NNF

FF

N

NNF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

oo

oo



Solution

Resultant Force

( ) ( )N

NNFR

629

8.5828.236 22

=

+=

From vector addition,

Direction angle θ is

N629=

o9.67

8.236

8.582tan 1

=

= −

N

Nθ



Solution

Cartesian Vector Notation

F1 = { 600cos30°i + 600sin30°j } N

F2 = { -400sin45°i + 400cos45°j } NF2 = { -400sin45°i + 400cos45°j } N

Thus,

FR = F1 + F2

= (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j

= {236.8i + 582.8j}N



Example 2.7

The end of the boom O is subjected to three

concurrent and coplanar forces. Determine

the magnitude and orientation of the the magnitude and orientation of the

resultant force.



Solution

Scalar Notation

Σ= FF xRx

4

:

↑=

+=

Σ=

←=−=

−+−=

N

NNF

FF

NN

NNNF

Ry

yRy

Rx

8.296

5

320045cos250

:

2.3832.383

5

420045sin250400

o

o

Solution

Resultant Force



( ) ( )NNFR 8.2962.383 22 +−=

From vector addition,

Direction angle θ is

N485=

o8.37

2.383

8.296tan 1

=

= −

N

Nθ

6161103 2.4 addition of a system

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